Hindawi Publishing Corporation Chinese Journal of Mathematics Volume 2013, Article ID 107145, 7 pages http://dx.doi.org/10.1155/2013/107145

Research Article Incomplete 𝑘-Fibonacci and 𝑘-Lucas Numbers

José L. Ramírez

Instituto de Matematicas´ y sus Aplicaciones, Universidad Sergio Arboleda, Colombia

Correspondence should be addressed to JoseL.Ram´ ´ırez; [email protected]

Received 10 July 2013; Accepted 11 September 2013

Academic Editors: B. Li and W. van der Hoek

Copyright © 2013 JoseL.Ram´ ´ırez. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We define the incomplete k-Fibonacci and k-Lucas numbers; we study the recurrence relations and some properties of these numbers.

1. Introduction In [6],theauthorsdefinetheincompleteFibonacciandLucas 𝑝-numbers. Also the authors define the incomplete bivariate Fibonacci numbers and their generalizations have many Fibonacci and Lucas 𝑝-polynomials in [7]. interesting properties and applications to almost every field On the other hand, many kinds of generalizations of 𝐹 of science and art (e.g., see [1]). Fibonacci numbers 𝑛 are Fibonaccinumbershavebeenpresentedintheliterature.In defined by the recurrence relation particular, a generalization is the 𝑘-Fibonacci Numbers. For any positive real number 𝑘,the𝑘-Fibonacci sequence, 𝐹0 =0, 𝐹1 =1, 𝐹𝑛+1 =𝐹𝑛 +𝐹𝑛−1,𝑛⩾1. (1) say {𝐹𝑘,𝑛}𝑛∈N, is defined recurrently by There exist a lot of properties about Fibonacci numbers. 𝐹 =0, 𝐹 =1, 𝐹 =𝑘𝐹 +𝐹 ,𝑛⩾1. In particular, there is a beautiful combinatorial identity to 𝑘,0 𝑘,1 𝑘,𝑛+1 𝑘,𝑛 𝑘,𝑛−1 (4) Fibonacci numbers [1] In [8], 𝑘-Fibonacci numbers were found by studying ⌊(𝑛−1)/2⌋ 𝑛−𝑖−1 𝐹 = ∑ ( ). the recursive application of two geometrical transformations 𝑛 𝑖 (2) 𝑖=0 used in the four-triangle longest-edge (4TLE) partition. These numbers have been studied in several papers; see8 [ – From (2), Filipponi [2] introduced the incomplete Fibonacci 14]. 𝐹 (𝑠) 𝐿 (𝑠) numbers 𝑛 and the incomplete Lucas numbers 𝑛 .They For any positive real number 𝑘,the𝑘-Lucas sequence, say are defined by {𝐿 } 𝑘,𝑛 𝑛∈N, is defined recurrently by 𝑠 𝑛−1−𝑗 𝑛−1 𝐿 =2, 𝐿 =𝑘, 𝐿 =𝑘𝐿 +𝐿 . 𝐹 (𝑠) = ∑ ( ) (𝑛=1,2,3,...;0≤𝑠≤⌊ ⌋) , 𝑘,0 𝑘,1 𝑘,𝑛+1 𝑘,𝑛 𝑘,𝑛−1 (5) 𝑛 𝑗 2 𝑗=0 If 𝑘=1,wehavetheclassicalLucasnumbers.Moreover, 𝐿 =𝐹 +𝐹 ,𝑛⩾1 𝑠 𝑛 𝑛−𝑗 𝑛 𝑘,𝑛 𝑘,𝑛−1 𝑘,𝑛+1 ;see[15]. 𝐿 (𝑠) = ∑ ( ) (𝑛=1,2,3,...;0≤𝑠≤⌊ ⌋) . 𝑘 𝑛 𝑗 In [12], the explicit formula to -Fibonacci numbers is 𝑗=0 𝑛−𝑗 2 ⌊(𝑛−1)/2⌋ 𝑛−𝑖−1 (3) 𝐹 = ∑ ( )𝑘𝑛−2𝑖−1, 𝑘,𝑛 𝑖 (6) 𝑖=0 Further in [3], generating functions of the incomplete Fibonacci and Lucas numbers are determined. In [4], Djord- and the explicit formula of 𝑘-Lucas numbers is jevic´ gave the incomplete generalized Fibonacci and Lucas ⌊𝑛/2⌋ 𝑛 𝑛−𝑖 numbers. In [5], Djordjevic´ and Srivastava defined incom- 𝐿 (𝑥) = ∑ ( )𝑘𝑛−2i. 𝑘,𝑛 𝑖 (7) plete generalized Jacobsthal and Jacobsthal-Lucas numbers. 𝑖=0 𝑛−𝑖 2 Chinese Journal of Mathematics

From (6)and(7), we introduce the incomplete 𝑘- Proof. Use Definition 1 to rewrite the right-hand side of (11) Fibonacci and 𝑘-Lucas numbers and we obtain new recurrent as relations, new identities, and their generating functions. 𝑙+1 𝑛−𝑖 𝑙 𝑛−𝑖−1 𝑘∑ ( )𝑘𝑛−2𝑖 + ∑ ( )𝑘𝑛−2𝑖−1 𝑖 𝑖 𝑖=0 𝑖=0 2. The Incomplete 𝑘-Fibonacci Numbers 𝑙+1 𝑛−𝑖 𝑙+1 𝑛−𝑖 = ∑ ( )𝑘𝑛−2𝑖+1 + ∑ ( )𝑘𝑛−2𝑖+1 Definition 1. The incomplete 𝑘-Fibonacci numbers are 𝑖 𝑖−1 defined by 𝑖=0 𝑖=1 𝑙+1 𝑛−𝑖 𝑛−𝑖 𝑛 (13) =𝑘𝑛−2𝑖+1 (∑ [( )+( )]) − 𝑘𝑛+1 ( ) 𝑖 𝑖−1 −1 𝑙 𝑖=0 𝑙 𝑛−1−𝑖 𝑛−2𝑖−1 𝑛−1 𝐹 = ∑ ( )𝑘 , 0≤𝑙≤⌊ ⌋. (8) 𝑘,𝑛 𝑖 2 𝑙+1 𝑖=0 𝑛−𝑖+1 = ∑ ( )𝑘𝑛−2𝑖+1 −0 𝑖 𝑖=0 𝑘 𝑙 In Table 1, some values of incomplete -Fibonacci num- =𝐹𝑘,𝑛+2. bers are provided. We note that

Proposition 3. One has 𝐹⌊(𝑛−1)/2⌋ =𝐹. 1,𝑛 𝑛 (9) 𝑠 𝑠 𝑛−𝑠−1 ∑ ( ) 𝐹𝑙+𝑖 𝑘𝑖 =𝐹𝑙+𝑠 , 0≤𝑙≤ . 𝑖 𝑘,𝑛+𝑖 𝑘,𝑛+2𝑠 (14) 𝑖=0 2 𝑘=1 For , we get incomplete Fibonacci numbers [2]. Proof (by induction on 𝑠). Sum (14)clearlyholdsfor𝑠=0and Some special cases of (8)are 𝑠=1(see (11)). Now suppose that the result is true for all 𝑗<𝑠+1;weproveitfor𝑠+1:

𝐹0 =𝑘𝑛−1; (𝑛≥1) , 𝑠+1 𝑠+1 𝑘,𝑛 ∑ ( )𝐹𝑙+𝑖 𝑘𝑖 𝑖 𝑘,𝑛+𝑖 1 𝑛−1 𝑛−3 𝑖=0 𝐹𝑘,𝑛 =𝑘 + (𝑛−2) 𝑘 ; (𝑛≥3) 𝑠+1 𝑠 𝑠 𝑙+𝑖 𝑖 2 𝑛−1 𝑛−3 (𝑛−4)(𝑛−3) 𝑛−5 = ∑ [( )+( )] 𝐹 𝑘 𝐹 =𝑘 + (𝑛−2) 𝑘 + 𝑘 ; (𝑛≥5) 𝑖 𝑖−1 𝑘,𝑛+𝑖 𝑘,𝑛 2 𝑖=0

𝐹⌊(𝑛−1)/2⌋ =𝐹 ; (𝑛≥1) 𝑠+1 𝑠 𝑠+1 𝑠 𝑘,𝑛 𝑘,𝑛 = ∑ ( )𝐹𝑙+𝑖 𝑘𝑖 + ∑ ( )𝐹𝑙+𝑖 𝑘𝑖 𝑖 𝑘,𝑛+𝑖 𝑖−1 𝑘,𝑛+𝑖 𝑛𝑘 𝑖=0 𝑖=0 ⌊(𝑛−3)/2⌋ {𝐹 − (𝑛 ) 𝐹 = 𝑘,𝑛 even (𝑛≥3) . 𝑠 𝑘,𝑛 { 2 𝑠 𝑠 𝐹 −1 (𝑛 ) =𝐹𝑙+𝑠 +( )𝐹𝑙+𝑠+1 𝑘𝑠+1 + ∑ ( )𝐹𝑙+𝑖+1 𝑘𝑖+1 { 𝑘,𝑛 odd 𝑘,𝑛+2𝑠 𝑠+1 𝑘,𝑛+𝑠+1 𝑖 𝑘,𝑛+𝑖+1 (10) 𝑖=−1 𝑠 𝑠 𝑠 =𝐹𝑙+𝑠 +0+∑ ( )𝐹𝑙+𝑖+1 𝑘𝑖+1 +( )𝐹𝑙 𝑘,𝑛+2𝑠 𝑖 𝑘,𝑛+𝑖+1 −1 𝑘,𝑛 𝑙 𝑖=0 2.1. Some Recurrence Properties of the Numbers 𝐹𝑘,𝑛 Proposition 2. 𝑘 𝑠 The recurrence relation of the incomplete - 𝑙+𝑠 𝑠 𝑙+𝑖+1 𝑖 𝑙 =𝐹𝑘,𝑛+2𝑠 +𝑘∑ ( )𝐹𝑘,𝑛+𝑖+1𝑘 +0 Fibonacci numbers 𝐹𝑘,𝑛 is 𝑖 𝑖=0

𝑙+𝑠 𝑙+𝑠+1 =𝐹𝑘,𝑛+2𝑠 +𝑘𝐹𝑘,𝑛+2𝑠+1 𝑙+1 𝑙+1 𝑙 𝑛−2 𝐹𝑘,𝑛+2 =𝑘𝐹𝑘,𝑛+1 +𝐹𝑘,𝑛,0≤𝑙≤ . (11) 𝑙+𝑠+1 2 =𝐹𝑘,𝑛+2𝑠+2. (15)

The relation (11) canbetransformedintothenonhomoge- neous recurrence relation Proposition 4. For 𝑛≥2𝑙+2,

𝑠−1 𝑙 𝑙 𝑙 𝑛−1−𝑙 𝑛−1−2𝑙 ∑𝐹𝑙 𝑘𝑠−1−𝑖 =𝐹𝑙+1 −𝑘𝑠𝐹𝑙+1 . 𝐹𝑘,𝑛+2 =𝑘𝐹𝑘,𝑛+1 +𝐹𝑘,𝑛 −( )𝑘 . (12) 𝑘,𝑛+𝑖 𝑘,𝑛+𝑠+1 𝑘,𝑛+1 (16) 𝑙 𝑖=0 Chinese Journal of Mathematics 3

𝑙 Table 1: The numbers 𝐹𝑘,𝑛,for1⩽𝑛⩽10. n\l 01 2 3 4 11 2 k 2 3 k2 𝑘 +1 3 4 k3 𝑘 +2𝑘 4 2 4 2 5 k4 𝑘 +3𝑘 𝑘 +3𝑘 +1 5 3 5 3 6 k5 𝑘 +4𝑘 𝑘 +4𝑘 +3𝑘 6 4 6 4 2 6 4 2 7 k6 𝑘 +5𝑘 𝑘 +5𝑘 +6𝑘 𝑘 +5𝑘 +6𝑘 +1 7 5 7 5 3 7 5 3 8 k7 𝑘 +6𝑘 𝑘 +6𝑘 + 10𝑘 𝑘 +6𝑘 + 10𝑘 +4𝑘 8 6 8 6 4 8 6 4 2 8 6 4 2 9 k8 𝑘 +7𝑘 𝑘 +7𝑘 + 15𝑘 𝑘 +7𝑘 + 15𝑘 + 10𝑘 𝑘 +7𝑘 + 15𝑘 + 10𝑘 +1 9 7 9 7 5 9 7 5 3 9 7 5 3 10 k9 𝑘 +8𝑘 𝑘 +8𝑘 + 21𝑘 𝑘 +8𝑘 + 21𝑘 + 20𝑘 𝑘 +8𝑘 + 21𝑘 + 20𝑘 +5𝑘

Proof(byinductionon𝑠). Sum (16) clearly holds for 𝑠=1 By deriving into the previous equation (respect to 𝑘), it is (see (11)). Now suppose that the result is true for all 𝑗<𝑠. obtained We prove it for 𝑠:

𝑠 𝑠−1 𝑙 𝑠−𝑖 𝑙 𝑠−𝑖−1 𝑙 ∑𝐹𝑘,𝑛+𝑖𝑘 =𝑘∑𝐹𝑘,𝑛+𝑖𝑘 +𝐹𝑘,𝑛+𝑠 ⌊(𝑛−1)/2⌋ 𝑛−1−𝑖 𝑖=0 𝑖=0 𝐹 +𝑘𝐹󸀠 = ∑ (𝑛−2𝑖) ( )𝑘𝑛−2𝑖−1 𝑘,𝑛 𝑘,𝑛 𝑖 =𝑘(𝐹𝑙+1 −𝑘𝑠𝐹𝑙+1 )+𝐹𝑙 𝑖=0 𝑘,𝑛+𝑠+1 𝑘,𝑛+1 𝑘,𝑛+𝑠 (17) (22) 𝑙+1 𝑙 𝑠+1 𝑙+1 =(𝑘𝐹 +𝐹 )−𝑘 𝐹 ⌊(𝑛−1)/2⌋ 𝑘,𝑛+𝑠+1 𝑘,𝑛+𝑠 𝑘,𝑛+1 𝑛−1−𝑖 =𝑛𝐹 −2 ∑ 𝑖( )𝑘𝑛−2𝑖−1. 𝑘,𝑛 𝑖 𝑙+1 𝑠+1 𝑙+1 𝑖=0 =𝐹𝑘,𝑛+𝑠+2 −𝑘 𝐹𝑘,𝑛+1.

Note that if 𝑘,in(4), is a real variable, then 𝐹𝑘,𝑛 =𝐹𝑥,𝑛 and From Lemma 5, they correspond to the Fibonacci polynomials defined by 1 𝑛=0 { if 𝐹 (𝑥) = 𝑥 𝑛=1 𝑛+1 { if (18) 𝑛𝐿 −𝑘𝐹 𝑥𝐹 (𝑥) +𝐹 (𝑥) 𝑛>1. 𝐹 +𝑘( 𝑘,𝑛 𝑘,𝑛 ) { 𝑛 𝑛−1 if 𝑘,𝑛 𝑘2 +4 Lemma 5. One has (23) ⌊(𝑛−1)/2⌋ 󸀠 𝑛𝐹𝑛+1 (𝑥) −𝑥𝐹𝑛 (𝑥) +𝑛𝐹𝑛−1 (𝑥) 𝑛−1−𝑖 𝐹 (𝑥) = =𝑛𝐹 −2 ∑ 𝑖( )𝑘𝑛−2𝑖−1. 𝑛 𝑥2 +4 𝑘,𝑛 𝑖 (19) 𝑖=0 𝑛𝐿 (𝑥) −𝑥𝐹 (𝑥) = 𝑛 𝑛 . 𝑥2 +4 See Proposition 13 of [12]. From where, after some algebra (20)isobtained. Lemma 6. One has Proposition 7. One has ⌊(𝑛−1)/2⌋ 2 𝑛−1−𝑖 ((𝑘 +4)𝑛−4)𝐹𝑘,𝑛 −𝑛𝑘𝐿𝑘,𝑛 ∑ 𝑖( )𝑘𝑛−1−2𝑖 = . 𝑖 2 𝑖=0 2(𝑘 +4) (20) 4𝐹 +𝑛𝑘𝐿 { 𝑘,𝑛 𝑘,𝑛 { (𝑛 even) Proof. From (6)wehavethat ⌊(𝑛−1)/2⌋ { 2(𝑘2 +4) 𝑙 { ∑ 𝐹𝑘,𝑛 = { 2 (24) ⌊(𝑛−1)/2⌋ 𝑛−1−𝑖 {(𝑘 +8)𝐹 +𝑛𝑘𝐿 𝑘𝐹 = ∑ ( ) 𝑘𝑛−2𝑖. 𝑙=0 { 𝑘,𝑛 𝑘,𝑛 (𝑛 ) . 𝑘,𝑛 𝑖 (21) 2 odd 𝑖=0 { 2(𝑘 +4) 4 Chinese Journal of Mathematics

𝑙 Table 2: The numbers 𝐿𝑘,𝑛,for1⩽𝑛⩽9. n\l 01 2 3 4 1 k 2 2 2 𝑘 𝑘 +2 3 3 3 𝑘 𝑘 +3𝑘 4 4 2 4 2 4 𝑘 𝑘 +4𝑘 𝑘 +4𝑘 +2 5 5 3 5 3 5 𝑘 𝑘 +5𝑘 𝑘 +5𝑘 +5𝑘 6 6 4 6 4 2 6 4 2 6 𝑘 𝑘 +6𝑘 𝑘 +6𝑘 +9𝑘 𝑘 +6𝑘 +9𝑘 +2 7 7 5 7 5 3 7 5 3 7 𝑘 𝑘 +7𝑘 𝑘 +7𝑘 + 14𝑘 𝑘 +7𝑘 + 14𝑘 +7𝑘 8 8 6 8 6 4 8 6 4 2 8 6 4 2 8 𝑘 𝑘 +8𝑘 𝑘 +8𝑘 + 20𝑘 𝑘 +8𝑘 + 20𝑘 + 16𝑘 𝑘 +8𝑘 + 20𝑘 + 16𝑘 +2 9 9 7 9 7 5 9 7 5 3 9 7 5 3 9 𝑘 𝑘 +9𝑘 𝑘 +9𝑘 + 27𝑘 𝑘 +9𝑘 + 27𝑘 + 30𝑘 𝑘 +9𝑘 + 27𝑘 + 30𝑘 +9𝑘

⌊(𝑛−1)/2⌋ Proof 𝑛−1 𝑛−1−𝑖 = ∑ (⌊ ⌋+1)( )𝑘𝑛−1−2𝑖 𝑖 ⌊(𝑛−1)/2⌋ 𝑖=0 2 𝑙 ∑ 𝐹𝑘,𝑛 𝑙=0 ⌊(𝑛−1)/2⌋ 𝑛−1−𝑖 − ∑ 𝑖( )𝑘𝑛−1−2𝑖 𝑖 0 1 ⌊(𝑛−1)/2⌋ 𝑖=0 =𝐹𝑘,𝑛 +𝐹𝑘,𝑛 +⋅⋅⋅+𝐹𝑘,𝑛

⌊(𝑛−1)/2⌋ 𝑛−1−0 𝑛−1 𝑛−1−𝑖 𝑛−1−2𝑖 =( )𝑘𝑛−1 =(⌊ ⌋+1)𝐹 − ∑ 𝑖( )𝑘 . 0 𝑘,𝑛 𝑖 2 𝑖=0 𝑛−1−0 𝑛−1−1 (25) +[( )𝑘𝑛−1 +( )𝑘𝑛−3] 0 1 From Lemma 6,(24)isobtained.

[ 𝑘 [ 𝑛−1−0 𝑛−1−1 3. The Incomplete -Lucas Numbers +⋅⋅⋅+[ ( )𝑘𝑛−1 +( )𝑘𝑛−3 [ 0 1 Definition 8. The incomplete 𝑘-Lucas numbers are defined by [ 𝑙 𝑛 𝑛−𝑖 𝑛 𝐿𝑙 = ∑ ( )𝑘𝑛−2𝑖, 0≤𝑙≤⌊ ⌋. 𝑛−1 𝑘,𝑛 𝑛−𝑖 𝑖 2 (26) 𝑛−1−⌊ ⌋ 𝑖=0 2 ] 𝑛−1−2⌊(𝑛−1)/2⌋] +⋅⋅⋅+( 𝑛−1 )𝑘 ] In Table 2,somenumbersofincomplete𝑘-Lucas numbers ⌊ ⌋ ] 2 are provided. ] We note that 𝑛−1 𝑛−1−0 𝐿⌊𝑛/2⌋ =𝐿 . =(⌊ ⌋+1)( )𝑘𝑛−1 1,𝑛 𝑛 (27) 2 0 Some special cases of (26)are 𝑛−1 𝑛−1−1 +⌊ ⌋( )𝑘𝑛−3 𝐿0 =𝑘𝑛; (𝑛≥1) , 2 1 𝑘,𝑛 𝐿1 =𝑘𝑛 +𝑛𝑘𝑛−2; (𝑛≥2) , 𝑛−1 𝑘,𝑛 𝑛−1−⌊ ⌋ 2 𝑛 (𝑛−3) 𝑛−1−2⌊(𝑛−1)/2⌋ 𝐿2 =𝑘𝑛 +𝑛𝑘𝑛−2 + 𝑘𝑛−4; (𝑛≥4) , +⋅⋅⋅+( 𝑛−1 )𝑘 𝑘,𝑛 2 ⌊ ⌋ (28) 2 ⌊𝑛/2⌋ 𝐿𝑘,𝑛 =𝐿𝑘,𝑛; (𝑛≥1) ,

⌊(𝑛−1)/2⌋ 𝑛−1 𝑛−1−𝑖 𝐿𝑘,𝑛 −2 (𝑛 even) = ∑ (⌊ ⌋+1−𝑖)( )𝑘𝑛−1−2𝑖 𝐿⌊(𝑛−2)/2⌋ ={ (𝑛≥2) . 𝑖 𝑘,𝑛 𝑖=0 2 𝐿𝑘,𝑛 −𝑛𝑘 (𝑛 odd) Chinese Journal of Mathematics 5

𝑙 3.1. Some Recurrence Properties of the Numbers 𝐿𝑘,𝑛 Proposition 12. One has

Proposition 9. One has 𝑠 𝑠 𝑛−𝑠 ∑ ( )𝐿𝑙+𝑖 𝑘𝑖 =𝐿𝑙+𝑠 ,0≤𝑙≤ . 𝑛 𝑖 𝑘,𝑛+𝑖 𝑘,𝑛+2𝑠 2 (37) 𝐿𝑙 =𝐹𝑙−1 +𝐹𝑙 ,0≤𝑙≤⌊⌋. 𝑖=0 𝑘,𝑛 𝑘,𝑛−1 𝑘,𝑛+1 2 (29) Proof. Using (29)and(14), we write Proof. By (8), rewrite the right-hand side of (29)as 𝑠 𝑠 𝑙−1 𝑙 𝑠 𝑠 𝑛−2−𝑖 𝑛−𝑖 ∑ ( )𝐿𝑙+𝑖 𝑘𝑖 = ∑ ( )[𝐹𝑙+𝑖−1 +𝐹𝑙+𝑖 ]𝑘𝑖 ∑ ( )𝑘𝑛−2−2𝑖 + ∑ ( )𝑘𝑛−2𝑖 𝑖 𝑘,𝑛+𝑖 𝑖 𝑘,𝑛+𝑖−1 𝑘,𝑛+𝑖+1 𝑖 𝑖 𝑖=0 𝑖=0 𝑖=0 𝑖=0 𝑠 𝑠 𝑠 𝑠 𝑙 𝑛−1−𝑖 𝑙 𝑛−𝑖 = ∑ ( )𝐹𝑙+𝑖−1 𝑘𝑖 + ∑ ( )𝐹𝑙+𝑖 𝑘𝑖 (38) = ∑ ( )𝑘𝑛−2𝑖 + ∑ ( )𝑘𝑛−2𝑖 𝑖 𝑘,𝑛+𝑖−1 𝑖 𝑘,𝑛+𝑖+1 𝑖−1 𝑖 𝑖=0 𝑖=0 𝑖=1 𝑖=0 =𝐹𝑙−1+𝑠 +𝐹𝑙+𝑠 =𝐿𝑙+𝑠 . 𝑙 𝑛−1−𝑖 𝑛−𝑖 𝑛−1 (30) 𝑘,𝑛−1+2𝑠 𝑘,𝑛+1+2𝑠 𝑘,𝑛+2𝑠 = ∑ [( )+( )] 𝑘𝑛−2𝑖 −( ) 𝑖−1 𝑖 −1 𝑖=0

𝑙 𝑛 𝑛−𝑖 Proposition 13. For 𝑛≥2𝑙+1, = ∑ ( )𝑘𝑛−2𝑖 +0 𝑛−𝑖 𝑖 𝑖=0 𝑠−1 𝑙 𝑠−1−𝑖 𝑙+1 𝑠 𝑙+1 𝑙 ∑𝐿𝑘,𝑛+𝑖𝑘 =𝐿𝑘,𝑛+𝑠+1 −𝑘𝐿𝑘,𝑛+1. (39) =𝐿𝑘,𝑛. 𝑖=0

The proof can be done by using (31) and induction on 𝑠. Proposition 10. The recurrence relation of the incomplete 𝑘- Lemma 14. 𝑙 One has Lucas numbers 𝐿𝑘,𝑛 is ⌊𝑛/2⌋ 𝑛 𝑛 𝑛−𝑖 𝑛−2𝑖 𝑛 𝑙+1 𝑙+1 𝑙 ∑ 𝑖 ( )𝑘 = [𝐿𝑘,𝑛 −𝑘𝐹𝑘,𝑛]. (40) 𝐿𝑘,𝑛+2 =𝑘𝐿𝑘,𝑛+1 +𝐿𝑘,𝑛, 0≤𝑙≤⌊ ⌋. (31) 𝑛−𝑖 𝑖 2 2 𝑖=0

The relation (31) canbetransformedintothenonhomoge- The proof is similar to Lemma 6. neous recurrence relation 𝑛 𝑛−𝑙 Proposition 15. One has 𝐿𝑙 =𝑘𝐿𝑙 +𝐿𝑙 − ( )𝑘𝑛−2𝑙. 𝑘,𝑛+2 𝑘,𝑛+1 𝑘,𝑛 𝑙 (32) 𝑛−𝑙 𝑛𝑘 ⌊𝑛/2⌋ { {𝐿𝑘,𝑛 + 𝐹𝑘,𝑛 (𝑛 even) Proof. Using (29)and(11), we write ∑ 𝐿𝑙 = 2 𝑘,𝑛 {1 (41) 𝑙=0 (𝐿 +𝑛𝑘𝐹 ) (𝑛 ) . 𝑙+1 𝑙 𝑙+1 {2 𝑘,𝑛 𝑘,𝑛 odd 𝐿𝑘,𝑛+2 =𝐹𝑘,𝑛+1 +𝐹𝑘,𝑛+3 𝑙 𝑙−1 𝑙+1 𝑙 Proof. An argument analogous to that of the proof of =𝑘𝐹𝑘,𝑛 +𝐹𝑘,𝑛−1 +𝑘𝐹𝑘,𝑛+2 +𝐹𝑘,𝑛+1 (33) Proposition 7 yields 𝑙 𝑙+1 𝑙−1 𝑙 =𝑘(𝐹𝑘,𝑛 +𝐹𝑘,𝑛+2)+𝐹𝑘,𝑛−1 +𝐹𝑘,𝑛+1 ⌊𝑛/2⌋ 𝑛 ⌊𝑛/2⌋ 𝑛 𝑛−𝑖 ∑ 𝐿𝑙 =(⌊ ⌋+1)𝐿 − ∑ 𝑖 ( )𝑘𝑛−2𝑖. =𝑘𝐿𝑙+1 +𝐿𝑙 . 𝑘,𝑛 2 𝑘,𝑛 𝑛−𝑖 𝑖 (42) 𝑘,𝑛+1 𝑘,𝑛 𝑙=0 𝑖=0

From Lemma 14,(41)isobtained. Proposition 11. One has 𝑛−1 4. Generating Functions of the Incomplete 𝑘𝐿𝑙 =𝐹𝑙 −𝐹𝑙−2 , 0≤𝑙≤⌊ ⌋. 𝑘 𝑘 𝑘,𝑛 𝑘,𝑛+2 𝑘,𝑛−2 2 (34) -Fibonacci and -Lucas Number Proof. By (29), In this section, we give the generating functions of incomplete 𝑘-Fibonacci and 𝑘-Lucas numbers. 𝐹𝑙 =𝐿𝑙 −𝐹𝑙−1,𝐹𝑙−2 =𝐿𝑙−1 −𝐹𝑙−1, 𝑘,𝑛+2 𝑘,𝑛+1 𝑘,𝑛 𝑘,𝑛−2 𝑘,𝑛−1 𝑘,𝑛 (35) ∞ Lemma 16 (see [3,page592]).Let {𝑠𝑛}𝑛=0 be a complex whence, from (31), sequence satisfying the following nonhomogeneous recurrence 𝑙 𝑙−2 𝑙 𝑙−1 𝑙 relation: 𝐹𝑘,𝑛+2 −𝐹𝑘,𝑛−2 =𝐿𝑘,𝑛+1 −𝐿𝑘,𝑛−1 =𝑘𝐿𝑘,𝑛. (36)

𝑠𝑛 =𝑎𝑠𝑛−1 +𝑏𝑠𝑛−2 +𝑟𝑛 (𝑛>1) , (43) 6 Chinese Journal of Mathematics

where 𝑎 and 𝑏 are complex numbers and {𝑟𝑛} is a given complex Now let sequence. Then, the generating function 𝑈(𝑡) of the sequence 𝑠 =𝐿𝑙 ,𝑠=𝐿𝑙 ,𝑠=𝐿𝑙 . {𝑠𝑛} is 0 𝑘,2𝑙 1 𝑘,2𝑙+1 𝑛 𝑘,𝑛+2𝑙 (51) 𝐺 (𝑡) +𝑠 −𝑟 +(𝑠 −𝑠 𝑎−𝑟)𝑡 Also let 𝑈 (𝑡) = 0 0 1 0 1 , (44) 1−𝑎𝑡−𝑏𝑡2 𝑛+2𝑙−2 𝑟 =𝑟 =0, 𝑟 =( )𝑘𝑛+2𝑙−2. 0 1 𝑛 𝑛+𝑙−2 (52) where 𝐺(𝑡) denotes the generating function of {𝑟𝑛}. 2 The generating function of the sequence {𝑟𝑛} is 𝐺(𝑡) =𝑡 (2 − Theorem 17. The generating function of the incomplete 𝑘- 𝑙+1 𝑙 𝑡)/(1 − 𝑘𝑡) (see [16,page355]).Thus,fromLemma 16,we Fibonacci numbers 𝐹𝑘,𝑛 is given by get the generating function 𝑆𝑘,𝑙(𝑥) of sequence {𝑠𝑛}. ∞ 𝑙 𝑖 𝑅𝑘,𝑙 (𝑥) = ∑𝐹𝑘,𝑖𝑡 𝑖=0 5. Conclusion 2 In this paper, we introduce incomplete 𝑘-Fibonacci and 2𝑙+1 𝑡 =𝑡 [𝐹𝑘,2𝑙+1 +(𝐹𝑘,2𝑙+2 −𝑘𝐹𝑘,2𝑙+1)𝑡− ] 𝑘-Lucas numbers, and we obtain new identities. In [17], (1−𝑘𝑡)𝑙+1 the authors introduced the ℎ(𝑥)-Fibonacci polynomials. −1 That generalizes Catalan’s Fibonacci polynomials and the ×[1−𝑘𝑡−𝑡2] . 𝑘-Fibonacci numbers. Let ℎ(𝑥) be a polynomial with real (45) coefficients. The ℎ(𝑥)-Fibonacci polynomials {𝐹ℎ,𝑛(𝑥)}𝑛∈N are defined by the recurrence relation Proof. Let 𝑙 be a fixed positive integer. From (8)and(12), 𝑙 𝑙 𝑙 𝐹𝑘,𝑛 =0for 0 ≤ 𝑛 < 2𝑙+1, 𝐹𝑘,2𝑙+1 =𝐹𝑘,2𝑙+1,and𝐹𝑘,2𝑙+2 = 𝐹ℎ,0 (𝑥) =0, 𝐹ℎ,1 (𝑥) =1, 𝐹𝑘,2𝑙+2,and (53) 𝐹ℎ,𝑛+1 (𝑥) =ℎ(𝑥) 𝐹ℎ,𝑛 (𝑥) +𝐹ℎ,𝑛−1 (𝑥) ,𝑛⩾1. 𝑛−3−𝑙 𝐹𝑙 =𝑘𝐹𝑙 +𝐹𝑙 −( )𝑘𝑛−3−2𝑙. 𝑘,𝑛 𝑘,𝑛−1 𝑘,𝑛−2 𝑙 (46) It would be interesting to study a definition of incomplete ℎ(𝑥)-Fibonacci polynomials and research their properties. Now let 𝑙 𝑙 𝑙 Acknowledgments 𝑠0 =𝐹𝑘,2𝑙+1,𝑠1 =𝐹𝑘,2𝑙+2,𝑠𝑛 =𝐹𝑘,𝑛+2𝑙+1. (47) The author would like to thank the anonymous referees for Also let their helpful comments. The author was partially supported 𝑛+𝑙−2 𝑛−2 by Universidad Sergio Arboleda under Grant no. USA-II- 𝑟0 =𝑟1 =0, 𝑟𝑛 =( )𝑘 . (48) 𝑛−2 2011-0059.

2 The generating function of the sequence {𝑟𝑛} is 𝐺(𝑡) =𝑡 /(1 − 𝑙+1 References 𝑘𝑡) (see [16,page355]).Thus,fromLemma 16,wegetthe generating function 𝑅𝑘,𝑙(𝑥) of sequence {𝑠𝑛}. [1] T. Koshy, Fibonacci and Lucas Numbers with Applications, Wiley-Interscience, 2001. Theorem 18. The generating function of the incomplete 𝑘- [2] P.Filipponi, “Incomplete Fibonacci and Lucas numbers,” Rendi- 𝑙 Lucas numbers 𝐿𝑘,𝑛 is given by conti del Circolo Matematico di Palermo,vol.45,no.1,pp.37–56, 1996. ∞ ´ 𝑆 (𝑥) = ∑𝐿𝑙 𝑡𝑖 [3] A. Pinter´ and H. M. Srivastava, “Generating functions of 𝑘,𝑙 𝑘,𝑖 the incomplete Fibonacci and Lucas numbers,” Rendiconti del 𝑖=0 Circolo Matematico di Palermo,vol.48,no.3,pp.591–596,1999. 𝑡2 (2−𝑡) [4] G. B. Djordjevic,´ “Generating functions of the incomplete gen- =𝑡2𝑙 [𝐿 +(𝐿 −𝑘𝐿 )𝑡− ] (49) 𝑘,2𝑙 𝑘,2𝑙+1 𝑘,2𝑙 𝑙+1 eralized Fibonacci and generalized Lucas numbers,” Fibonacci (1−𝑘𝑡) Quarterly,vol.42,no.2,pp.106–113,2004. 2 −1 [5] G. B. Djordjevic´ and H. M. Srivastava, “Incomplete generalized ×[1−𝑘𝑡−𝑡 ] . Jacobsthal and Jacobsthal-Lucas numbers,” Mathematical and Computer Modelling,vol.42,no.9-10,pp.1049–1056,2005. Proof. Theproofofthistheoremissimilartotheproofof [6] D. Tasci and M. Cetin Firengiz, “Incomplete Fibonacci and Theorem 17.Let𝑙 be a fixed positive integer. From (26)and 𝑙 𝑙 𝑙 Lucas p-numbers,” Mathematical and Computer Modelling,vol. (32), 𝐿𝑘,𝑛 =0for 0≤𝑛<2𝑙, 𝐿𝑘,2𝑙 =𝐿𝑘,2𝑙,and𝐿𝑘,2𝑙+1 =𝐿𝑘,2𝑙+1, 52, no. 9-10, pp. 1763–1770, 2010. and [7] D. Tasci, M. Cetin Firengiz, and N. Tuglu, “Incomplete bivariate Fibonaccu and Lucas p-polynomials,” Discrete Dynamics in 𝑙 𝑙 𝑙 𝑛−2 𝑛−2−𝑙 𝑛−2−2𝑙 𝐿 =𝑘𝐿 +𝐿 − ( )𝑘 . Nature and Society, vol. 2012, Article ID 840345, 11 pages, 2012. 𝑘,𝑛 𝑘,𝑛−1 𝑘,𝑛−2 𝑛−2−𝑙 𝑛−2−2𝑙 [8] S. Falcon´ and A.´ Plaza, “On the Fibonacci k-numbers,” Chaos, (50) Solitons and Fractals,vol.32,no.5,pp.1615–1624,2007. Chinese Journal of Mathematics 7

[9] C. Bolat and H. Kose,¨ “On the properties of k-Fibonacci numbers,” International Journal of Contemporary Mathematical Sciences, vol. 22, no. 5, pp. 1097–1105, 2010. [10] S. Falcon, “The k-Fibonacci matrix and the Pascal matrix,” Central European Journal of Mathematics,vol.9,no.6,pp.1403– 1410, 2011. [11] S. Falcon´ and A.´ Plaza, “The k-Fibonacci sequence and the Pascal 2-triangle,” Chaos, Solitons and Fractals,vol.33,no.1,pp. 38–49, 2007. [12] S. Falcon´ and A.´ Plaza, “On k-Fibonacci sequences and polyno- mials and their derivatives,” Chaos, Solitons and Fractals,vol.39, no. 3, pp. 1005–1019, 2009. [13] J. Ram´ırez, “Some properties of convolved k-Fibonacci num- bers,” ISRN Combinatorics,vol.2013,ArticleID759641,5pages, 2013. [14] A. Salas, “About k-Fibonacci numbers and their associated numbers,” International Mathematical Forum,vol.50,no.6,pp. 2473–2479, 2011. [15] S. Falcon, “On the k-Lucas numbers,” International Journal of Contemporary Mathematical Sciences,vol.21,no.6,pp.1039– 1050, 2011. [16] H. M. Srivastava and H. L. Manocha, A Treatise on Generating Functions, Halsted Press, Ellis Horwood Limited, Chichester, UK, John Wiley & Sons, New York, NY, USA, 1984. [17] A. Nalli and P. Haukkanen, “On generalized Fibonacci and Lucas polynomials,” Chaos, Solitons and Fractals,vol.42,no.5, pp.3179–3186,2009. Advances in Advances in Journal of Journal of Operations Research Decision Sciences Applied Mathematics Algebra Probability and Statistics Hindawi Publishing Corporation Hindawi Publishing Corporation Hindawi Publishing Corporation Hindawi Publishing Corporation Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014

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